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a) \(\sqrt{2x+2}-\sqrt{2x-1}=x\)
\(\Leftrightarrow2x+2+2x-1-2\sqrt{\left(2x+2\right)\left(2x-1\right)}=x^2\)
\(\Leftrightarrow4x+1-2\sqrt{\left(2x+2\right)\left(2x-1\right)}=x^2\)
\(\Leftrightarrow2\sqrt{4x^2+2x-2}=-x^2+4x+1\)( ĐK: \(2-\sqrt{5}\le x\le2+\sqrt{5}\))
\(\Leftrightarrow4\left(4x^2+2x-2\right)=\left(x^2-4x-1\right)^2\)
\(\Leftrightarrow16x^2+8x-8=x^4-8x^3+14x^2+8x+1\)
\(\Leftrightarrow x^4-8x^3-2x^2+9=0\)
\(\Leftrightarrow x^4-x^3-7x^3+7x^2-9x^2+9=0\)
\(\Leftrightarrow x^3\left(x-1\right)-7x^2\left(x-1\right)-9\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-7x^2-9x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(chon\right)\\x=8,22...\left(loai\right)\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất \(x=-1\)
b_em ko chắc đâu, chưa từng làm dạng toán chứa tham số-_-
ĐK: \(x^2\ge-m\) ( ko chắc)
PT<=> \(\left(x-3\right)\sqrt{x^2+m}=\left(x-3\right)\left(x+3\right)\)
\(\Leftrightarrow\left(x-3\right)\left[x+3-\sqrt{x^2+m}\right]=0\)
Thấy ngay x = 3 thỏa mãn. Xét cái ngoặc to
\(\Leftrightarrow x+3=\sqrt{x^2+m}\left(\text{thêm đk }x\ge-3\right)\Leftrightarrow6x+9=m\Leftrightarrow x=\frac{\left(m-9\right)}{6}\)
Do \(x\ge-3\text{nên }m\ge-9\)
Vậy...
Do tam giác ABC vuông tại A và \(\widehat{B}=30^o\) \(\Rightarrow C=60^o\)
\(\Rightarrow\left(\overrightarrow{AB},\overrightarrow{BC}\right)=150^o;\)\(\left(\overrightarrow{BA},\overrightarrow{BC}\right)=30^o;\left(\overrightarrow{AC},\overrightarrow{CB}\right)=120^o\)
\(\left(\overrightarrow{AB},\overrightarrow{AC}\right)=90^o;\left(\overrightarrow{BC},\overrightarrow{BA}\right)=30^o\).Do vậy:
a) \(\cos\left(\overrightarrow{AB},\overrightarrow{BC}\right)+\sin\left(\overrightarrow{BA},\overrightarrow{BC}\right)+\tan\frac{\left(\overrightarrow{AC},\overrightarrow{CB}\right)}{2}\)
\(=\cos150^o+\sin30^o+\tan60^o\)
\(=-\frac{\sqrt{3}}{2}+\frac{1}{2}+\sqrt{3}\)
\(=\frac{\sqrt{3}+1}{2}\)
b) \(\sin\left(\overrightarrow{AB},\overrightarrow{AC}\right)+\cos\left(\overrightarrow{BC},\overrightarrow{AB}\right)+\cos\left(\overrightarrow{CA},\overrightarrow{BA}\right)\)
\(=\sin90^o+\cos30^o+\cos0^o\)
\(=1+\frac{\sqrt{3}}{2}\)
\(=\frac{2+\sqrt{3}}{2}\)
\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)
\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)
\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)
\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)
\(E=cot^2x\left(cos^2x-1\right)+cos^2x+2=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+2=2\)
\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)
Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)
\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)
\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)
1/ Tinh ∆. Pt co 2 nghiem x1,x2 <=> ∆>=0.
Theo dinh ly Viet: S=x1+x2=-b/a=m+3.
Theo gt: |x1|=|x2| <=> ...
2/ \(\frac{\sin^2x-\cos^2x}{1+2\sin x.\cos x}\)
\(=\frac{\cos^2x\left(\frac{\sin^2x}{\cos^2x}-\frac{\cos^2x}{\cos^2x}\right)}{\cos^2x\left(\frac{1}{\cos^2x}+\frac{2\sin x.\cos x}{\cos^2x}\right)}\)
\(=\frac{\tan^2x-1}{\tan^2x+1+2\tan x}\)
\(=\frac{\left(\tan x-1\right)\left(\tan x+1\right)}{\left(\tan x+1\right)^2}\)
\(=\frac{\tan x-1}{\tan x+1}\left(dpcm\right)\)
c/ A M C B N BC=8 AC=7 AB=6
\(\Leftrightarrow BA^2=CA^2-2\overrightarrow{CA}.\overrightarrow{CB}+CB^2\)
\(\Leftrightarrow\overrightarrow{CA}.\overrightarrow{CB}=\frac{CA^2+CB^2-BA^2}{2}=\frac{77}{2}\)
\(\Leftrightarrow MN^2=\frac{9}{4}CB^2-\frac{15}{7}\overrightarrow{CA}.\overrightarrow{CB}+\frac{25}{49}CA^2\)
\(=\frac{9}{4}.64-\frac{15}{7}.\frac{77}{2}+\frac{25}{49}.49\)
\(=\frac{173}{2}\)
\(\Rightarrow MN=\sqrt{\frac{173}{2}}=\frac{\sqrt{346}}{2}\)